If a variable line drawn through the intersection of the line `x/3+y/4=1&x/4+y/3=1`, meets the coordinate axes at A and B, `(A!=B)` , then the locus of the midpoint of AB is:

Options

A. 7x y = 6 ( x + y)
B. 6xy = 7(x + y)
C. `4 (x + y)^(2) - 28 (x + y) + 49 = 0`
D. `14 (x + y)^(2) - 97 (x + y) + 168 = 0`

Answer

A

Explanation

The equation of a line through the intersection of the lines `(x)/(3) + (y)/(4) = 1` and `(x)/(4) + (y)/(3) = 1` is
`((x )/(3) + (y)/(4) - 1) + lambda ((x)/(4) + (y)/(3) -1) = 0`
`implies x ((1)/(3) + (lambda)/(4)) + y ((1)/(4) + (lambda)/(3)) = lambda + 1`
This meets the coordinate axes at
`A ((lambda + 1) /(1//3 + lambda//4) , 0) and B (0 , (lambda + 1)/(1 //4 + lambda//3))` respectively .
Let (h,k) be the coordinates of the mid-point of AB . Then ,
`2h = (lambda + 1)/((1)/(3) + (lambda)/(4))` and `2k = (lambda + 1)/((1)/(4) + (lambda)/(3))`
`implies (1)/(2h) + (1)/(2k) = (((1)/(3) + (lambda)/(3)) + ((1)/(4) + (lambda)/(3)))/(lambda + 1) implies (h + k)/(2hk ) = (7)/(12) implies 6 (h + k) = 7hk`
Hence , the locus of (h,k) is 6 (x + y) = 7xy .